Volume of a solid of rotation, obtained rotating a function around x=2

[greg_rack]

Homework Statement

Given the parabola $y=4-x^2$, restricted to the first quadrant, compute the volume of the solid obtained by rotating the area enclosed by the parabola and xy axis around the line $x=2$

Relevant Equations

Definite integration

At first, I inverted the function($f^{-1}(x)=g(x)$) and calculated the volume through the integral:
$$V=\pi\int_{0}^{4}[4-(2-g(x))^2]\ dx$$
but then I questioned myself if the same result could have been obtained without inverting the function.

To find such a strategy, I proceeded as follows:
in order to get to a Riemann sum, I divided interval [a;b] into n small intervals $dx$; for each of those I took an arbitrary point $c_i$, and computed the volume of a single "slice" of final solid as $V_i=f(c_i)\cdot 2\pi c_i \cdot dx$.
The total volume is thus $V=2\pi \int_{a}^{b}xf(x) \ dx$, which applied to my case:
$$V=2\pi \int_{2}^{0}x(4-x^2) \ dx=-2\pi \int_{0}^{2}x(4-x^2) \ dx$$
but something doesn't work.

 

[etotheipi]

The first method looks okay!

As for the second, it's not obvious to me what the geometry of your slices $V_i$ are but I think you were trying to construct cylindrical shells centred on the axis of rotation $(2, t,0)$?

That should work, but note that the radii of these cylindrical shells are actually $r_i = 2 - c_i$, and not $c_i$. So the area of the annulus cross-section is $2\pi(2-c_i)dx$ and hence the volume of the shell is $V_i = 2\pi (2-c_i) f(c_i) dx$, leading to$$V = 2\pi \int_0^2 (2-x)(4-x^2) dx$$and that gives the same as your first result!

 

[greg_rack]

The first method looks okay!

As for the second, it's not obvious to me what the geometry of your slices $V_i$ are but I think you were trying to construct cylindrical shells centred on the axis of rotation $(2, t,0)$?

That should work, but note that the radii of these cylindrical shells are actually $r_i = 2 - c_i$, and not $c_i$. So the area of the annulus cross-section is $2\pi(2-c_i)dx$ and hence the volume of the shell is $V_i = 2\pi (2-c_i) f(c_i) dx$, leading to$$V = 2\pi \int_0^2 (2-x)(4-x^2) dx$$and that gives the same as your first result!

Brilliant! I didn't notice I had to switch from $c_i$ to $2-c_i$, and that makes perfectly sense :)